实验5

任务1：

 1 #include <stdio.h>
 2 #define N 5
 3 
 4 void input(int x[], int n);
 5 void output(int x[], int n);
 6 void find_min_max(int x[], int n, int *pmin, int *pmax);
 7 
 8 int main() {
 9     int a[N];
10     int min, max;
11 
12     printf("录入%d个数据:\n", N);
13     input(a, N);
14 
15     printf("数据是: \n");
16     output(a, N);
17 
18     printf("数据处理...\n");
19     find_min_max(a, N, &min, &max);
20 
21     printf("输出结果:\n");
22     printf("min = %d, max = %d\n", min, max);
23 
24     return 0;
25 }
26 
27 void input(int x[], int n) {
28     int i;
29 
30     for(i = 0; i < n; ++i)
31         scanf("%d", &x[i]);
32 }
33 
34 void output(int x[], int n) {
35     int i;
36     
37     for(i = 0; i < n; ++i)
38         printf("%d ", x[i]);
39     printf("\n");
40 }
41 
42 void find_min_max(int x[], int n, int *pmin, int *pmax) {
43     int i;
44     
45     *pmin = *pmax = x[0];
46 
47     for(i = 0; i < n; ++i)
48         if(x[i] < *pmin)
49             *pmin = x[i];
50         else if(x[i] > *pmax)
51             *pmax = x[i];
52 }

 问题1.找出最大值，最小值

问题2.指向x[0]的地址

 1 #include <stdio.h>
 2 #define N 5
 3 
 4 void input(int x[], int n);
 5 void output(int x[], int n);
 6 int *find_max(int x[], int n);
 7 
 8 int main() {
 9     int a[N];
10     int *pmax;
11 
12     printf("录入%d个数据:\n", N);
13     input(a, N);
14 
15     printf("数据是: \n");
16     output(a, N);
17 
18     printf("数据处理...\n");
19     pmax = find_max(a, N);
20 
21     printf("输出结果:\n");
22     printf("max = %d\n", *pmax);
23 
24     return 0;
25 }
26 
27 void input(int x[], int n) {
28     int i;
29 
30     for(i = 0; i < n; ++i)
31         scanf("%d", &x[i]);
32 }
33 
34 void output(int x[], int n) {
35     int i;
36     
37     for(i = 0; i < n; ++i)
38         printf("%d ", x[i]);
39     printf("\n");
40 }
41 
42 int *find_max(int x[], int n) {
43     int max_index = 0;
44     int i;
45 
46     for(i = 0; i < n; ++i)
47         if(x[i] > x[max_index])
48             max_index = i;
49     
50     return &x[max_index];
51 }

 问题1：最大值的指针

问题2：可以

任务2：

 1 #include <stdio.h>
 2 #include <string.h>
 3 #define N 80
 4 
 5 int main() {
 6     char s1[N] = "Learning makes me happy";
 7     char s2[N] = "Learning makes me sleepy";
 8     char tmp[N];
 9 
10     printf("sizeof(s1) vs. strlen(s1): \n");
11     printf("sizeof(s1) = %d\n", sizeof(s1));
12     printf("strlen(s1) = %d\n", strlen(s1));
13 
14     printf("\nbefore swap: \n");
15     printf("s1: %s\n", s1);
16     printf("s2: %s\n", s2);
17 
18     printf("\nswapping...\n");
19     strcpy(tmp, s1);
20     strcpy(s1, s2);
21     strcpy(s2, tmp);
22 
23     printf("\nafter swap: \n");
24     printf("s1: %s\n", s1);
25     printf("s2: %s\n", s2);
26 
27     return 0;
28 }

 问题1：数组s1大小为80个字节，sizeof(s1)计算的是数组s1所占字节， strlen(s1)统计的是字符串长度（不包括"\0")

问题2：不能，s1代表的是地址，不能输入字符串

问题3：交换

 1 #include <stdio.h>
 2 #include <string.h>
 3 #define N 80
 4 
 5 int main() {
 6     char *s1 = "Learning makes me happy";
 7     char *s2 = "Learning makes me sleepy";
 8     char *tmp;
 9 
10     printf("sizeof(s1) vs. strlen(s1): \n");
11     printf("sizeof(s1) = %d\n", sizeof(s1));
12     printf("strlen(s1) = %d\n", strlen(s1));
13 
14     printf("\nbefore swap: \n");
15     printf("s1: %s\n", s1);
16     printf("s2: %s\n", s2);
17 
18     printf("\nswapping...\n");
19     tmp = s1;
20     s1 = s2;
21     s2 = tmp;
22 
23     printf("\nafter swap: \n");
24     printf("s1: %s\n", s1);
25     printf("s2: %s\n", s2);
26 
27     return 0;
28 }

 问题1：指向L的地址， s1指针地址的所占字节， 字符串长度

问题2：能， task2_1.c中s1是地址常量，不能被赋值， task2_2.中s1是指针变量，可以被赋值

问题3：s1和s2的地址，没有

任务3：

 1 #include <stdio.h>
 2 
 3 int main()
 4 {
 5     int x[2][4] = {{1, 9, 8, 4}, {2, 0, 4, 9}};
 6     int i, j;
 7     int *ptr1;     // 指针变量，存放int类型数据的地址
 8     int(*ptr2)[4]; // 指针变量，指向包含4个int元素的一维数组
 9 
10     printf("输出1: 使用数组名、下标直接访问二维数组元素\n");
11     for (i = 0; i < 2; ++i)
12     {
13         for (j = 0; j < 4; ++j)
14             printf("%d ", x[i][j]);
15         printf("\n");
16     }
17 
18     printf("\n输出2: 使用指针变量ptr1(指向元素)间接访问\n");
19     for (ptr1 = &x[0][0], i = 0; ptr1 < &x[0][0] + 8; ++ptr1, ++i)
20     {
21         printf("%d ", *ptr1);
22 
23         if ((i + 1) % 4 == 0)
24             printf("\n");
25     }
26 
27     printf("\n输出3: 使用指针变量ptr2(指向一维数组)间接访问\n");
28     for (ptr2 = x; ptr2 < x + 2; ++ptr2)
29     {
30         for (j = 0; j < 4; ++j)
31             printf("%d ", *(*ptr2 + j));
32         printf("\n");
33     }
34 
35     return 0;
36 }

 问题：int (*ptr)[4]中ptr 是一个指针，它存储的是一个包含 4 个 int 元素的一维数组的地址。

 int *ptr[4]中ptr是一个包含 4 个指针的数组，每个指针可以独立地指向不同的 int 数据。

任务4：

 1 #include <stdio.h>
 2 #define N 80
 3 
 4 void replace(char *str, char old_char, char new_char);
 5 
 6 int main() {
 7     char text[N] = "Programming is difficult or not, it is a question.";
 8 
 9     printf("原始文本: \n");
10     printf("%s\n", text);
11 
12     replace(text, 'i', '*');
13 
14     printf("处理后文本: \n");
15     printf("%s\n", text);
16 
17     return 0;
18 }
19 
20 void replace(char *str, char old_char, char new_char) {
21     int i;
22 
23     while(*str) {
24         if(*str == old_char)
25             *str = new_char;
26         str++;
27     }
28 }

 问题1：将i替换成*

问题2：可以

任务5：

 1 #include <stdio.h>
 2 #define N 80
 3 
 4 char *str_trunc(char *str, char x);
 5 
 6 int main()
 7 {
 8     char str[N];
 9     char ch;
10 
11     while (printf("输入字符串: "), gets(str) != NULL)
12     {
13         printf("输入一个字符: ");
14         ch = getchar();
15 
16         printf("截断处理...\n");
17         str_trunc(str, ch);
18 
19         printf("截断处理后的字符串: %s\n\n", str);
20         getchar();
21     }
22 
23     return 0;
24 }
25 
26 char *str_trunc(char *str, char x)
27 {
28     char *p = str;
29     while (*p)
30     {
31         if (*p == x)
32         {
33             *p = '\0';
34         }
35         else
36             p++;
37     }
38     return str;
39 }

 会自动将换行认为是下一个字符串，使下一次输入直接跳到输入阻断字符

去除缓冲

任务6：

 1 #include <stdio.h>
 2 #include <string.h>
 3 #define N 5
 4 
 5 int check_id(char *str); // 函数声明
 6 
 7 int main()
 8 {
 9     char *pid[N] = {"31010120000721656X",
10                     "3301061996X0203301",
11                     "53010220051126571",
12                     "510104199211197977",
13                     "53010220051126133Y"};
14     int i;
15 
16     for (i = 0; i < N; ++i)
17         if (check_id(pid[i]))
18             printf("%s\tTrue\n", pid[i]);
19         else
20             printf("%s\tFalse\n", pid[i]);
21 
22     return 0;
23 }
24 
25 int check_id(char *str)
26 {
27     if (strlen(str) != 18)
28         return 0;
29     for (int i = 0; i < 17; i++)
30     {
31         if (str[i] < '0' || str[i] > '9')
32             return 0;
33     }
34     if (str[17] == 'X' || (str[17] >= '0' && str[17] <= '9'))
35         return 1;
36     return 0;
37 }

 任务7：

 1 #include <stdio.h>
 2 #define N 80
 3 void encoder(char *str, int n); // 函数声明
 4 void decoder(char *str, int n); // 函数声明
 5 
 6 int main()
 7 {
 8     char words[N];
 9     int n;
10 
11     printf("输入英文文本: ");
12     gets(words);
13 
14     printf("输入n: ");
15     scanf("%d", &n);
16 
17     printf("编码后的英文文本: ");
18     encoder(words, n); // 函数调用
19     printf("%s\n", words);
20 
21     printf("对编码后的英文文本解码: ");
22     decoder(words, n); // 函数调用
23     printf("%s\n", words);
24 
25     return 0;
26 }
27 
28 /*函数定义
29 功能：对s指向的字符串进行编码处理
30 编码规则：
31 对于a~z或A~Z之间的字母字符，用其后第n个字符替换; 其它非字母字符，保持不变
32 */
33 void encoder(char *str, int n)
34 {
35     for (int i = 0; str[i] != '\0'; i++)
36     {
37         if (str[i] >= 'a' && str[i] <= 'z')
38         {
39             if (str[i] + n >= 'a' && str[i] + n <= 'z')
40             {
41                 str[i] = str[i] + n;
42             }
43             else
44             {
45                 str[i] = str[i] + n - 26;
46             }
47         }
48         if (str[i] >= 'A' && str[i] <= 'Z')
49         {
50             if (str[i] + n >= 'A' && str[i] + n <= 'Z')
51             {
52                 str[i] = str[i] + n;
53             }
54             else
55             {
56                 str[i] = str[i] + n - 26;
57             }
58         }
59     }
60 }
61 
62 /*函数定义
63 功能：对s指向的字符串进行解码处理
64 解码规则：
65 对于a~z或A~Z之间的字母字符，用其前面第n个字符替换; 其它非字母字符，保持不变
66 */
67 void decoder(char *str, int n)
68 {
69     for (int i = 0; str[i] != '\0'; i++)
70     {
71         if (str[i] >= 'a' && str[i] <= 'z')
72         {
73             if (str[i] - n >= 'a' && str[i] - n <= 'z')
74             {
75                 str[i] = str[i] - n;
76             }
77             else
78             {
79                 str[i] = str[i] - n + 26;
80             }
81         }
82         if (str[i] >= 'A' && str[i] <= 'Z')
83         {
84             if (str[i] - n >= 'A' && str[i] - n <= 'Z')
85             {
86                 str[i] = str[i] - n;
87             }
88             else
89             {
90                 str[i] = str[i] - n + 26;
91             }
92         }
93     }
94 }

 任务8：

 1 #include <stdio.h>
 2 #include <string.h>
 3 
 4 void sort(char *argv[], int left, int right);
 5 int partition(char *argv[], int left, int right);
 6 void swap(char **a, char **b);
 7 
 8 int main(int argc, char *argv[])
 9 {
10     sort(argv, 1, argc - 1);
11     int i;
12     for ( i = 1; i < argc; ++i)
13         printf("hello, %s\n", argv[i]);
14 
15     return 0;
16 }
17 
18 void sort(char *argv[], int left, int right)
19 {
20     if (left < right)
21     {
22         int pivotIndex = partition(argv, left, right);
23         sort(argv, left, pivotIndex - 1);
24         sort(argv, pivotIndex + 1, right);
25     }
26 }
27 
28 int partition(char *argv[], int left, int right)
29 {
30     char *pivot = argv[right];
31     int i = left - 1;
32     int j;
33     for ( j = left; j < right; j++)
34     {
35         if (strcmp(argv[j], pivot) <= 0)
36         {
37             i++;
38             swap(&argv[i], &argv[j]);
39         }
40     }
41     swap(&argv[i + 1], &argv[right]);
42     return i + 1;
43 }
44 
45 void swap(char **a, char **b)
46 {
47     char *temp = *a;
48     *a = *b;
49     *b = temp;
50 }

 

posted @ 2024-12-02 16:54  Erhjiu  阅读(14)  评论(0)  编辑  收藏  举报